Questão 13 - 1ª Fase - IME 2025

 

 

Pelo enunciado, tem-se:

$\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}=2$

Pelo teorema da bissetriz interna:

$\frac{\overline{\mathrm{BD}}}{\overline{\mathrm{DC}}}=\frac{\mathrm{c}}{\mathrm{b}}$ e $\overline{\mathrm{BD}}+\overline{\mathrm{DC}}=\mathrm{a}\Rightarrow \overline{\mathrm{BD}}=\frac{\mathrm{ac}}{\mathrm{b}+\mathrm{c}}=\frac{\mathrm{c}}{2}$ e $\overline{\mathrm{DC}} \frac{\mathrm{ba}}{\mathrm{b}+\mathrm{c}}=\frac{\mathrm{b}}{2}$

Da figura:

$$\begin{gathered} \mathrm{B}\hat{\mathrm{A}}\mathrm{L}=\mathrm{C}\hat{\mathrm{A}}\mathrm{L}=\mathrm{\alpha } \\ \mathrm{A}\hat{\mathrm{C}}\mathrm{B}=\mathrm{A}\hat{\mathrm{L}}\mathrm{B}=\mathrm{\gamma }=\frac{\widehat{\mathrm{AB}}}{2} \end{gathered}$$

Os triângulos ADC e ABL são semelhantes pelo caso AA, de maneira que:

$\frac{\overline{\mathrm{BL}}}{\overline{\mathrm{AL}}}=\frac{\overline{\mathrm{DC}}}{\overline{\mathrm{AC}}}=\frac{{\displaystyle \frac{\mathrm{b}}{2}}}{\mathrm{b}}=\boxed{\frac{1}{2}}$